Discrete Structures - CM0246Lattices

Andrés Sicard-Ramírez

Universidad EAFIT

Semester 2014-2

Lattices from the Partial Orders Theory

Definition (Lattice)A lattice (retículo) is a poset where every pair of elements has both asupremum and an infimum.

ExampleThe following poset is a lattice.

𝑎

𝑏𝑐

𝑑

𝑒

𝑓

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Lattices from the Partial Orders Theory

Definition (Lattice)A lattice (retículo) is a poset where every pair of elements has both asupremum and an infimum.

ExampleThe following poset is a lattice.

𝑎

𝑏𝑐

𝑑

𝑒

𝑓

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Lattices from the Partial Orders Theory

Example (counter-example)The following poset is not a lattice because the upper bounds of the pair{𝑏, 𝑐} are 𝑑, 𝑒 and 𝑓 , but this set has not a least upper bound.

𝑎

𝑏 𝑐

𝑑 𝑒

𝑓

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Lattices from the Partial Orders Theory

Example (counter-example)The following poset is not a lattice because for example, the pair {1, 2}has not supremum.

⊥

2 …10 𝑛 …

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Lattices from the Partial Orders Theory

Examples(ℤ+, ∣) is a lattice where the supremum is the least common multipleand the infimum is the greatest common divisor.

Let 𝐴 be a set. Is (𝑃 (𝐴), ⊆) a lattice?

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Lattices from the Partial Orders Theory

Examples(ℤ+, ∣) is a lattice where the supremum is the least common multipleand the infimum is the greatest common divisor.Let 𝐴 be a set. Is (𝑃 (𝐴), ⊆) a lattice?

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Algebraic Structures

Definition (Algebraic structure)An algebraic structure on a set 𝐴 ≠ 0 is essentially a collection of 𝑛-aryoperations on 𝐴 (Cohn 1981, p. 41).

Example (Semigroup)A semigroup (𝑆, ∗) is a set 𝑆 with an associative binary operation∗ ∶ 𝑆 × 𝑆 → 𝑆.

Example (Monoid)A monoid (𝑀, ∗, 𝜀) is a semigroup (𝑀, ∗) with an element 𝜀 ∈ 𝑀 which isan unit for ∗, i.e. (∀𝑥)(𝑥 ∗ 𝜀 = 𝜀 ∗ 𝑥 = 𝑥).

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Lattices from the Algebraic Structures Theory

Definition (Lattice)Let ∧ and ∨ be two binaries operations, called meet and join, respectively.A lattice (retículo) is an algebraic structure (𝐿, ∧, ∨), which satisfy thefollowing axioms for all 𝑥, 𝑦 and 𝑧 in 𝐿 (Lipschutz and Lipson 2007):

𝑥 ∧ 𝑦 = 𝑦 ∧ 𝑥 (Commutative laws)𝑥 ∨ 𝑦 = 𝑦 ∨ 𝑥

(𝑥 ∧ 𝑦) ∧ 𝑧 = 𝑥 ∧ (𝑦 ∧ 𝑧) (Associative laws)(𝑥 ∨ 𝑦) ∨ 𝑧 = 𝑥 ∨ (𝑦 ∨ 𝑧)

𝑥 ∧ (𝑥 ∨ 𝑦) = 𝑥 (Absortion laws)𝑥 ∨ (𝑥 ∧ 𝑦) = 𝑥

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Lattices from the Algebraic Structures Theory

ExampleLet 𝐴 be a set. (𝑃 (𝐴), ∩, ∪) is a lattice.

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Lattices from the Algebraic Structures Theory

Definition (Dual)The dual of any statement in a lattice (𝐿, ∧, ∨) is the statement obtainedby interchanging ∧ and ∨.

ExampleThe dual of 𝑥 ∧ (𝑦 ∨ 𝑥) = 𝑥 ∨ 𝑥 is 𝑥 ∨ (𝑦 ∧ 𝑥) = 𝑥 ∧ 𝑥.

Theorem (Principle of duality)The dual of any theorem in a lattice is also an theorem (Lipschutz andLipson 2007).

Proof.The dual of every axiom in a lattice is also an axiom. Hence, the dualtheorem can be proved by using the dual of each step of the proof of theoriginal theorem.

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Lattices from the Algebraic Structures Theory

Definition (Dual)The dual of any statement in a lattice (𝐿, ∧, ∨) is the statement obtainedby interchanging ∧ and ∨.

ExampleThe dual of 𝑥 ∧ (𝑦 ∨ 𝑥) = 𝑥 ∨ 𝑥 is 𝑥 ∨ (𝑦 ∧ 𝑥) = 𝑥 ∧ 𝑥.

Theorem (Principle of duality)The dual of any theorem in a lattice is also an theorem (Lipschutz andLipson 2007).

Proof.The dual of every axiom in a lattice is also an axiom. Hence, the dualtheorem can be proved by using the dual of each step of the proof of theoriginal theorem.

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Lattices from the Algebraic Structures Theory

Definition (Dual)The dual of any statement in a lattice (𝐿, ∧, ∨) is the statement obtainedby interchanging ∧ and ∨.

ExampleThe dual of 𝑥 ∧ (𝑦 ∨ 𝑥) = 𝑥 ∨ 𝑥 is 𝑥 ∨ (𝑦 ∧ 𝑥) = 𝑥 ∧ 𝑥.

Theorem (Principle of duality)The dual of any theorem in a lattice is also an theorem (Lipschutz andLipson 2007).

Proof.The dual of every axiom in a lattice is also an axiom. Hence, the dualtheorem can be proved by using the dual of each step of the proof of theoriginal theorem.

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Lattices from the Algebraic Structures Theory

Definition (Dual)The dual of any statement in a lattice (𝐿, ∧, ∨) is the statement obtainedby interchanging ∧ and ∨.

ExampleThe dual of 𝑥 ∧ (𝑦 ∨ 𝑥) = 𝑥 ∨ 𝑥 is 𝑥 ∨ (𝑦 ∧ 𝑥) = 𝑥 ∧ 𝑥.

Theorem (Principle of duality)The dual of any theorem in a lattice is also an theorem (Lipschutz andLipson 2007).

Proof.The dual of every axiom in a lattice is also an axiom. Hence, the dualtheorem can be proved by using the dual of each step of the proof of theoriginal theorem.

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Lattices from the Algebraic Structures Theory

ExampleLet (𝐿, ∧, ∨) be a lattice. Prove the idempotent laws

𝑥 ∧ 𝑥 = 𝑥, (1)𝑥 ∨ 𝑥 = 𝑥. (2)

Proof of (1).𝑥 ∧ 𝑥 = 𝑥 ∧ (𝑥 ∨ (𝑥 ∧ 𝑦)) (second absortion law)

= 𝑥 (first absortion law)

Proof of (2).By principle of duality on (1).

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Lattices from the Algebraic Structures Theory

ExampleLet (𝐿, ∧, ∨) be a lattice. Prove the idempotent laws

𝑥 ∧ 𝑥 = 𝑥, (1)𝑥 ∨ 𝑥 = 𝑥. (2)

Proof of (1).𝑥 ∧ 𝑥 = 𝑥 ∧ (𝑥 ∨ (𝑥 ∧ 𝑦)) (second absortion law)

= 𝑥 (first absortion law)

Proof of (2).By principle of duality on (1).

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Lattices from the Algebraic Structures Theory

ExampleLet (𝐿, ∧, ∨) be a lattice. Prove the idempotent laws

𝑥 ∧ 𝑥 = 𝑥, (1)𝑥 ∨ 𝑥 = 𝑥. (2)

Proof of (1).𝑥 ∧ 𝑥 = 𝑥 ∧ (𝑥 ∨ (𝑥 ∧ 𝑦)) (second absortion law)

= 𝑥 (first absortion law)

Proof of (2).By principle of duality on (1).

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Lattices from the Algebraic Structures Theory

Exercise (Rosen (2004), Problem 40, p. 500)Prove that if 𝑥 and 𝑦 are elements of a lattice (𝐿, ∧, ∨) then 𝑥 ∨ 𝑦 = 𝑦, ifand only if, 𝑥 ∧ 𝑦 = 𝑥.

Proof →.Let’s suppose 𝑥 ∨ 𝑦 = 𝑦. Then

𝑥 = 𝑥 ∧ (𝑥 ∨ 𝑦) (first absortion law)= 𝑥 ∧ 𝑦 (hypothesis)

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Lattices from the Algebraic Structures Theory

Exercise (Rosen (2004), Problem 40, p. 500)Prove that if 𝑥 and 𝑦 are elements of a lattice (𝐿, ∧, ∨) then 𝑥 ∨ 𝑦 = 𝑦, ifand only if, 𝑥 ∧ 𝑦 = 𝑥.

Proof →.Let’s suppose 𝑥 ∨ 𝑦 = 𝑦. Then

𝑥 = 𝑥 ∧ (𝑥 ∨ 𝑦) (first absortion law)= 𝑥 ∧ 𝑦 (hypothesis)

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Lattices from the Algebraic Structures Theory

Exercise (cont.)

Proof ←.Let’s suppose 𝑥 ∧ 𝑦 = 𝑥. Then

𝑦 = 𝑦 ∨ (𝑦 ∧ 𝑥) (second absortion law)= 𝑦 ∨ (𝑥 ∧ 𝑦) (commutative law)= 𝑦 ∨ 𝑥 (hypothesis)= 𝑥 ∨ 𝑦 (commutative law)

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Equivalence of the Definitions

TheoremLet (𝐿, ∧, ∨) be a lattice. Then (𝐿, ≼) is a partial order, where the relation ≼is defined by (Lipschutz and Lipson 2007):

𝑥 ≼ 𝑦 def= 𝑥 ∧ 𝑦 = 𝑥.

Proof.1. The relation ≼ is reflexive

𝑥 ∧ 𝑥 = 𝑥 (idempotency), for all 𝑥 ∈ 𝐿. Therefore 𝑥 ≼ 𝑥, for all𝑥 ∈ 𝐿.

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Equivalence of the Definitions

TheoremLet (𝐿, ∧, ∨) be a lattice. Then (𝐿, ≼) is a partial order, where the relation ≼is defined by (Lipschutz and Lipson 2007):

𝑥 ≼ 𝑦 def= 𝑥 ∧ 𝑦 = 𝑥.Proof.

1. The relation ≼ is reflexive𝑥 ∧ 𝑥 = 𝑥 (idempotency), for all 𝑥 ∈ 𝐿. Therefore 𝑥 ≼ 𝑥, for all𝑥 ∈ 𝐿.

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Equivalence of the Definitions

Proof (cont.)2. The relation ≼ is antisymmetric

Suppose 𝑥 ≼ 𝑦 and 𝑦 ≼ 𝑥, then 𝑥 ∧ 𝑦 = 𝑥 and 𝑦 ∧ 𝑥 = 𝑦. Therefore

𝑥 = 𝑥 ∧ 𝑦 (hypothesis)= 𝑦 ∧ 𝑥 (commutative law)= 𝑦 (hypothesis)

That is, ≼ is antisymmetric.

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Equivalence of the Definitions

Proof (cont.)3. The relation ≼ is transitive

Suppose 𝑥 ≼ 𝑦 and 𝑦 ≼ 𝑧, then 𝑥 ∧ 𝑦 = 𝑥 and 𝑦 ∧ 𝑧 = 𝑦. Therefore

𝑥 ∧ 𝑧 = (𝑥 ∧ 𝑦) ∧ 𝑧 (hypothesis)= 𝑥 ∧ (𝑦 ∧ 𝑧) (associativity law)= 𝑥 ∧ 𝑦 (hypothesis)= 𝑥 (hypothesis)

That is, 𝑥 ≼ 𝑧.

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Equivalence of the Definitions

RemarkLet (𝐿, ∧, ∨) be a lattice and let be (𝐿, ≼) the order partial induced by(𝐿, ∧, ∨). It is possible prove that (𝐿, ≼) is a lattice.

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Equivalence of the Definitions

Theorem (Rosen (2004), Problem 39, p. 500)Let (𝐿, ≼) be a lattice. Then (𝐿, ∧, ∨) is a lattice, where

𝑥 ∧ 𝑦 def= inf(𝑥, 𝑦),𝑥 ∨ 𝑦 def= sup(𝑥, 𝑦),

Proof.1. Commutative laws for ∧ and ∨ (Rosen’s solution).

Because inf(𝑥, 𝑦) = inf(𝑦, 𝑥) and sup(𝑥, 𝑦) = sup(𝑦, 𝑥), it followsthat 𝑥 ∧ 𝑦 = 𝑦 ∧ 𝑥 and 𝑥 ∨ 𝑦 = 𝑦 ∨ 𝑥.

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Equivalence of the Definitions

Theorem (Rosen (2004), Problem 39, p. 500)Let (𝐿, ≼) be a lattice. Then (𝐿, ∧, ∨) is a lattice, where

𝑥 ∧ 𝑦 def= inf(𝑥, 𝑦),𝑥 ∨ 𝑦 def= sup(𝑥, 𝑦),

Proof.1. Commutative laws for ∧ and ∨ (Rosen’s solution).

Because inf(𝑥, 𝑦) = inf(𝑦, 𝑥) and sup(𝑥, 𝑦) = sup(𝑦, 𝑥), it followsthat 𝑥 ∧ 𝑦 = 𝑦 ∧ 𝑥 and 𝑥 ∨ 𝑦 = 𝑦 ∨ 𝑥.

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Equivalence of the Definitions

Proof (cont.)2. Associative laws for ∧ and ∨ (Rosen’s solution).

Using the definition, (𝑥 ∧ 𝑦) ∧ 𝑧 is a lower bound of 𝑥, 𝑦 and 𝑧 thatis greater than every other lower bound. Because 𝑥, 𝑦 and 𝑧 playinterchangeable roles, 𝑥 ∧ (𝑦 ∧ 𝑧) is the same element.

Similarly, (𝑥 ∨ 𝑦) ∨ 𝑧 is an upper bound of 𝑥, 𝑦 and 𝑧 that is less thanevery other upper bound. Because 𝑥, 𝑦 and 𝑧 play interchangeableroles, 𝑥 ∨ (𝑦 ∨ 𝑧) is the same element.

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Equivalence of the Definitions

Proof (cont.)2. Associative laws for ∧ and ∨ (Rosen’s solution).

Using the definition, (𝑥 ∧ 𝑦) ∧ 𝑧 is a lower bound of 𝑥, 𝑦 and 𝑧 thatis greater than every other lower bound. Because 𝑥, 𝑦 and 𝑧 playinterchangeable roles, 𝑥 ∧ (𝑦 ∧ 𝑧) is the same element.

Similarly, (𝑥 ∨ 𝑦) ∨ 𝑧 is an upper bound of 𝑥, 𝑦 and 𝑧 that is less thanevery other upper bound. Because 𝑥, 𝑦 and 𝑧 play interchangeableroles, 𝑥 ∨ (𝑦 ∨ 𝑧) is the same element.

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Equivalence of the Definitions

Proof (cont.)3. Absortion laws for ∧ and ∨ (Rosen’s solution).

To show that 𝑥 ∧ (𝑥 ∨ 𝑦) = 𝑥 it is sufficient to show that 𝑥 is thegreatest lower bound of 𝑥, and 𝑥 ∨ 𝑦. Note that 𝑥 is a lower boundof 𝑥, and because 𝑥 ∨ 𝑦 is by definition greater than 𝑥, 𝑥 is a lowerbound for it as well. Therefore, 𝑥 is a lower bound. But any lowerbound of 𝑥 has to be less than 𝑥, so 𝑥 is the greatest lower bound.

The second statement is the dual of the first; we omit its proof.

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References

Cohn, P. M. (1981). Universal Algebra. Revised edition. D. ReidelPublishing Company (cit. on p. 8).Lipschutz, S. and Lipson, M. L. (2007). Schaum’s Outline of DiscreteMathematics. 3rd ed. McGraw-Hill (cit. on pp. 9, 11–14, 21, 22).Rosen, K. H. (2004). Matemática Discreta y sus Aplicaciones. 5th ed.McGraw-Hill (cit. on pp. 18, 19, 26, 27).

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